Find the equation of the normal line to the function y=3x+sin3x at point (0,0)

Answer 1

y = 3x + sin(3x)

y' = 3 + 3cos(3x)

Slope of tangent line at (0,0) is 3 + 3cos0 = 6

Slope of normal line (line perpendicular to the tangent line at (0,0) ) is -1/6

Equation of normal line: y - 0 = -(1/6)(x-0) So, y = -(1/6)x.